For someone with zero experience, this can be very confusing.
How do I define the Set Packing problem as seen in the wikipedia article as a MathProg program, to be later ran in the GLPK tool?
Intution alone would lead me into something like this:
var x
maximize SetPacking :
sum {s in Subsets} x
s.t. ?? //x is an integer 0 or 1
s.t. ?? //amount of x <=1
end;
But its logic is obviously wrong and I can't even finish it.
You can formulate the set packing problem in AMPL (and MathProg which is based on a subset of AMPL) as follows:
set U; # universe
set S; # subsets
set Members{S}; # members of subsets
# every set is either in the set packing or not
var x{S} binary;
# maximize the total number of subsets
maximize NumSets: sum{s in S} x[s];
# selected sets have to be pairwise disjoint
s.t. Disjoint{e in U}: sum{s in S: e in Members[s]} x[s] <= 1;
Note that you have to introduce indexed set Members with each element Members[s]representing members of subset s. This is the only difference from the algebraic formulation.
Related
Background: I have a simulation model which has unobserved parameters. I created a metamodel using artificial neural networks (ANN) because the runtime was very long for the simulation model. I am trying to estimate the unobserved parameters using Bayesian calibration, where priors are based on current knowledge, and the likelihood of observing data is being estimated from the metamodel.
Query: I have two random variables X and Y for which I am trying to get the posterior distribution using STAN. The prior distribution of X is uniform, U(0,2). The prior for Y is also uniform, but it will always exceed X i.e., Y ~ U(X,2). Since Y is linked to X, how can I define the prior distribution for Y in STAN such that the constraint Y>X holds? I am new to STAN, so I would appreciate any suggestions or guidance on how to proceed. Thank you so much!
Stan's ordered vectors are what you need. Create an ordered vector of length 2 (I'll call it beta) in the parameters block, like this:
parameters {
ordered<lower=0,upper=2>[2] beta;
}
Ordered vectors are constrained such that each element is greater than the previous element. So beta[1] will be your estimate of X and beta[2] will be your estimate of Y.
(To make sure I understand your model correctly: you have two parameters, X and Y, and your only prior knowledge about them is that they both lie in [0, 2] and Y > X. X and Y describe some aspect of the distribution of your data - for example, maybe X is the mean of some other random variable Z, for which you have observations. Do I have that right?)
I believe Stan's priors are uniform by default, but you can make sure of this by specifying a prior for beta in the model block:
model {
beta ~ uniform(0, 2);
...
}
When we need to optimize a function on the positive real half-line, and we only have non-constraints optimization routines, we use y = exp(x), or y = x^2 to map to the real line and still optimize on the log or the (signed) square root of the variable.
Can we do something similar for linear constraints, of the form Ax = b where, for x a d-dimensional vector, A is a (N,n)-shaped matrix and b is a vector of length N, defining the constraints ?
While, as Ervin Kalvelaglan says this is not always a good idea, here is one way to do it.
Suppose we take the SVD of A, getting
A = U*S*V'
where if A is n x m
U is nxn orthogonal,
S is nxm, zero off the main diagonal,
V is mxm orthogonal
Computing the SVD is not a trivial computation.
We first zero out the elements of S which we think are non-zero just due to noise -- which can be a slightly delicate thing to do.
Then we can find one solution x~ to
A*x = b
as
x~ = V*pinv(S)*U'*b
(where pinv(S) is the pseudo inverse of S, ie replace the non zero elements of the diagonal by their multiplicative inverses)
Note that x~ is a least squares solution to the constraints, so we need to check that it is close enough to being a real solution, ie that Ax~ is close enough to b -- another somewhat delicate thing. If x~ doesn't satisfy the constraints closely enough you should give up: if the constraints have no solution neither does the optimisation.
Any other solution to the constraints can be written
x = x~ + sum c[i]*V[i]
where the V[i] are the columns of V corresponding to entries of S that are (now) zero. Here the c[i] are arbitrary constants. So we can change variables to using the c[] in the optimisation, and the constraints will be automatically satisfied. However this change of variables could be somewhat irksome!
I'm learning AMPL to speed up a model currently in excel spreadsheet with excel solver. It basically based on the matrix multiplication result of a 1 x m variables and an m x n parameters. And it would find the variables to maximize the minimum of certain values in the result while keeping some other values in the same result satisfying a few constraints. How to do so in AMPL?
Given: P= m x n parameters
Variable: X= 1 x m variable we tried to solve
Calculate: R= X x P , result of matrix multiplication of X and P
Maximize: min(R[1..3]), the minimum value of the first 3 values in the result
Subject to: R[2]<R[4]
min(R[6..8])>20
R[5]-20>R[7]
X are all integers
I read several tutorials and look up the manual but can't find the solution to this seemingly straightforward problem. All I found is maximize a single value, which is the calculation result. And it was used only once and does not appear again in the constraint.
The usual approach for "maximize the minimum" problems in products like AMPL is to define an auxiliary variable and set linear constraints that effectively define it as the minimum, converting a nonlinear function (min) into linear rules.
For instance, suppose I have a bunch of decision variables x[i] with i ranging over an index set S, and I want to maximize the minimum over x[i]. AMPL syntax for that would be:
var x_min;
s.t. DefineMinimum{i in S}: x_min <= x[i];
maximize ObjectiveFunction: x_min;
The constraint only requires that x_min be less than or equal to the minimum of x[i]. However, since you're trying to maximize x_min and there are no other constraints on it, it should always end up exactly equal to that minimum (give or take machine-arithmetic epsilon considerations).
If you have parameters (i.e. values are known before you run the optimisation) and want to refer to their minimum, AMPL lets you do that more directly:
param p_min := min{j in IndexSet_P} p[j];
While AMPL also supports this syntax for variables, not all of the solvers used with AMPL are capable of accepting this type of constraint. For instance:
reset;
option solver gecode;
set S := {1,2,3};
var x{S} integer;
var x_min = min{s in S} x[s];
minimize OF: sum{s in S} x[s];
s.t. c1: x_min >= 5;
solve;
This will run and do what you'd expect it to do, because Gecode is programmed to recognise and deal with min-type constraints. However, if you switch the solver option to gurobi or cplex it will fail, since these only accept linear or quadratic constraints. To apply a minimum constraint with those solvers, you need to use something like the linearization trick I discussed above.
I have a problem which often occurs in energy system modelling (modelling of storages) where charging of a storage system is not allowed to when it is discharging and vice versa.
Let x and y be continous positive decision variables that represent the charging/discharging power of the storage unit. Now I am trying to express the following logic in equations or inequalities:
if x > 0:
y = 0
if y > 0:
x = 0
I know this can be realized with binary variables via the following equations if b is a binary variable and x_ub and y_ub represent the respective upper bound of the variables:
x <= b * x_ub
y <= (1-b) * y_ub
This results in a mixed-integer linear programm (MILP) and of course induces all the properties of a MILP which lay in higher runtimes, other interpretations of the solution, ...
Another way to get a pure LP formulation is to work with (very low) costs that are attached to x and y so that only one of both variables is set to a positive value. But this also affects my objective, has other disadvantages as is is somewhat of a hotfix.
So here's my question: Is there any way to express the logic above (if/else) in a purely linear program (LP)?
Thanks in advance!
How do you impose a constraint that all values in a vector you are trying to optimize for are greater than zero, using fmincon()?
According to the documentation, I need some parameters A and b, where A*x ≤ b, but I think if I make A a vector of -1's and b 0, then I will have optimized for the sum of x>0, instead of each value of x greater than 0.
Just in case you need it, here is my code. I am trying to optimize over a vector (x) such that the (componentwise) product of x and a matrix (called multiplierMatrix) makes a matrix for which the sum of the columns is x.
function [sse] = myfun(x) % this is a nested function
bigMatrix = repmat(x,1,120) .* multiplierMatrix;
answer = sum(bigMatrix,1)';
sse = sum((expectedAnswer - answer).^2);
end
xGuess = ones(1:120,1);
[sse xVals] = fmincon(#myfun,xGuess,???);
Let me know if I need to explain my problem better. Thanks for your help in advance!
You can use the lower bound:
xGuess = ones(120,1);
lb = zeros(120,1);
[sse xVals] = fmincon(#myfun,xGuess, [],[],[],[], lb);
note that xVals and sse should probably be swapped (if their name means anything).
The lower bound lb means that elements in your decision variable x will never fall below the corresponding element in lb, which is what you are after here.
The empties ([]) indicate you're not using linear constraints (e.g., A,b, Aeq,beq), only the lower bounds lb.
Some advice: fmincon is a pretty advanced function. You'd better memorize the documentation on it, and play with it for a few hours, using many different example problems.